Free falling body on Jupiter

FREE FALLING BODY ON JUPITER

Consider a body of mass m to be dropped at a height of 1kilometer above the surface of Jupiter. The body will experience a force of gravity equal to the acceleration due to gravity of Jupiter g = 25.9 m/s². This value can be considered to be approximately same 1kilometer above Jupiter’s surface. As the body falls toward Jupiter, the displacement is considered negative.

ASSUMPTIONS

·         1. Effect of air resistance on is negligible

·         2. Effect of wind is negligible

·         3. Effect of rotation of Jupiter is negligible

Initial Displacement y­’ = 0 m

Final Displacement y = -1000 m

Applying first kinematic equation we get,

y - y' = u*t - (1/2)g*t^2

u – Initial velocity m/s

-1000 = 0*t - (1/2)*25.9*t^2 

-1000 = -12.95*t^2

t^2 = 77.22

t = 8.7874 s

This is the time taken by the body to cover a distance of 1km.

The final velocity v when the body hits the surface is given by 3rd Kinematic equation,

v^2 = u^2 - 2g(y - y')

v^2 = 0 - 2*25.9*(-1000-0)

v^2 = 2*25.9*1000

                     

 v   = √2*25.9*1000

      

 v  = 227.59 m/s

Thus the final velocity of the body when it reaches surface of Jupiter is 227.59 m/s or 819.34kmph or 512.09mph